Question

A factory produces bulbs. The probability that one bulb is defective is \[\frac{1}{50}\] and they are packed in boxes of 10. From a single box, find the probability that  more than 8 bulbs work properly                                                                                                                            

 

 

Answer 1

Let getting a defective bulb from a single box is a success.
We have

\[p = \text{ probability of getting a defective bulb}  = \frac{1}{50}\]
\[\text{ Also,}  q = 1 - p = 1 - \frac{1}{50} = \frac{49}{50}\]
\[\text{ Let X denote the number of success in a sample of 10 trials . Then, } \]
\[\text{ X follows binomial distribution with parameters n = 10 and p }  = \frac{1}{50}\]
\[ \therefore P\left( X = r \right) = ^{10}{}{C}_r p^r q^\left( 10 - r \right) = ^{10}{}{C}_r \left( \frac{1}{50} \right)^r \left( \frac{49}{50} \right)^\left( 10 - r \right) = \frac{^{10}{}{C}_r {49}^\left( 10 - r \right)}{{50}^{10}}, \text{ where } r = 0, 1, 2, 3, . . . , 10\]
\[\text{ Now} , \]

\[ \text{ Required probability }  = P\left( \text{ more than 8 bulbs work properly } \right)\]
\[ = P\left(\text{  atmost one bulb is defective}  \right)\]
\[ = P\left( X \leq 0 \right)\]
\[ = P\left( X = 0 \right) + P\left( X = 1 \right)\]
\[ = \frac{^{10}{}{C}_0 {49}^{10}}{{50}^{10}} + \frac{^{10}{}{C}_1 {49}^9}{{50}^{10}}\]
\[ = \frac{{49}^{10}}{{50}^{10}} + \frac{10 \times {49}^9}{{50}^{10}}\]
\[ = \frac{{49}^9}{{50}^{10}}\left( 49 + 10 \right)\]
\[ = \frac{59\left( {49}^9 \right)}{\left( {50}^{10} \right)}\]